Câu 7: B
Câu 8: C
Câu 10: A
7.
ĐKXĐ: \(3\le x\le7\)
\(x-5+\sqrt{7-x}-\sqrt{x-3}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-3}=a\ge0\\\sqrt{7-x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=2\left(x-5\right)\)
Pt trở thành:
\(\dfrac{a^2-b^2}{2}+b-a=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=\sqrt{7-x}\\\sqrt{x-3}+\sqrt{7-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7-x\\4+2\sqrt{\left(x-3\right)\left(7-x\right)}=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\\x=7\end{matrix}\right.\)
8.
ĐKXĐ: \(-2\le x\le2\)
Đặt \(\sqrt{x+2}+\sqrt{2-x}=t\Rightarrow2\le t\le2\sqrt{2}\)
\(t^2=4+2\sqrt{4-x^2}\Rightarrow2\sqrt{-x^2+4}=t^2-4\)
Pt trở thành:
\(t+t^2-4+2m+3=0\)
\(\Leftrightarrow t^2+t+1=-2m\)
Xét hàm \(f\left(t\right)=t^2+t+1\) trên \(\left[2;2\sqrt{2}\right]\)
\(a=1>0;-\dfrac{b}{2a}=-\dfrac{1}{2}< 2\Rightarrow f\left(t\right)\) đồng biến trên đoạn đã cho
\(\Rightarrow f\left(2\right)\le f\left(t\right)\le f\left(2\sqrt{2}\right)\Rightarrow7\le f\left(t\right)\le9+2\sqrt{2}\)
\(\Rightarrow7\le-2m\le9+2\sqrt{2}\Rightarrow-\dfrac{9+2\sqrt{2}}{2}\le m\le-\dfrac{7}{2}\)
\(\Rightarrow m=\left\{-5;-4\right\}\)
10.
\(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)+1-25=0\)
\(\Leftrightarrow\left(x^2+x-1\right)^2-25=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Rightarrow x_1x_2=-6\) theo Viet
